Find the center and radius of the circle whose equation is x^2-4x+y^2+5y-17=0

To find the center and radius of the circle, we'll need to rewrite the given equation in the standard form of a circle's equation:

\[
(x - h)^2 + (y - k)^2 = r^2
\]

where \((h, k)\) is the center of the circle, and \(r\) is the radius. The given equation is:

\[
x^2 - 4x + y^2 + 5y - 17 = 0
\]

### Step 1: Rearrange and Complete the Square

1. Group the \(x\) and \(y\) terms:
   \[
   (x^2 - 4x) + (y^2 + 5y) = 17
   \]

2. Complete the square for the \(x\)-terms:
   - Take half of the coefficient of \(x\), which is \(-4\), divide by 2 to get \(-2\), and square it: \((-2)^2 = 4\).
   - Add and subtract 4 inside the equation.

   \[
   (x^2 - 4x + 4) - 4
   \]

   becomes

   \[
   (x - 2)^2 - 4
   \]

3. Complete the square for the \(y\)-terms:
   - Take half of the coefficient of \(y\), which is \(5\), divide by 2 to get \(\frac{5}{2}\), and square it: \(\left(\frac{5}{2}\right)^2 = \frac{25}{4}\).
   - Add and subtract \(\frac{25}{4}\) inside the equation.

   \[
   (y^2 + 5y + \frac{25}{4}) - \frac{25}{4}
   \]

   becomes

   \[
   (y + \frac{5}{2})^2 - \frac{25}{4}
   \]

### Step 2: Combine Everything Together

The equation now looks like:

\[
(x - 2)^2 - 4 + (y + \frac{5}{2})^2 - \frac{25}{4} = 17
\]

Combine the constants on the right side:

\[
(x - 2)^2 + (y + \frac{5}{2})^2 = 17 + 4 + \frac{25}{4}
\]

\[
(x - 2)^2 + (y + \frac{5}{2})^2 = \frac{68}{4} + \frac{25}{4}
\]

\[
(x - 2)^2 + (y + \frac{5}{2})^2 = \frac{93}{4}
\]

### Step 3: Identify the Center and Radius

The standard form of the equation now is:

\[
(x - 2)^2 + (y + \frac{5}{2})^2 = \left(\frac{\sqrt{93}}{2}\right)^2
\]

Thus:
- **Center**: \((h, k) = (2, -\frac{5}{2})\)
- **Radius**: \(r = \frac{\sqrt{93}}{2}\)

### Final Answer

- **Center**: \((2, -\frac{5}{2})\)
- **Radius**: \(\frac{\sqrt{93}}{2}\)

Would you like more details or have any questions about this process? 

Here are 5 related questions:
1. How do you complete the square for quadratic expressions?
2. What is the general form of a circle’s equation?
3. How can you derive the radius from the standard form of a circle?
4. Can a circle have a negative radius? Why or why not?
5. What happens if a circle equation has coefficients other than 1 for \(x^2\) and \(y^2\)?

**Tip:** Completing the square is a useful technique for converting quadratic expressions into a form that is easier to analyze or graph.

Learn how to find the center and radius of a circle from the equation x^2 - 4x + y^2 + 5y - 17 = 0. This step-by-step solution covers key geometry concepts such as completing the square and the standard form of a circle's equation. Suitable for students in Grades 9-12.

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